Problem: $\overline{AC}$ is $3$ units long $\overline{BC}$ is $10$ units long $\overline{AB}$ is $\sqrt{109}$ units long What is $\sin(\angle BAC)$ ? $A$ $C$ $B$ $3$ $10$ $\sqrt{109}$
Explanation: SOH CAH TOA in = pposite over ypotenuse opposite $= \overline{BC} = 10$ hypotenuse $= \overline{AB} = \sqrt{109}$ $\sin(\angle BAC)=\frac{10}{\sqrt{109}}$ $=\dfrac{10\sqrt{109} }{109}$